Is it possible to create a type alias to a generic class

Question

I would like to define a class type (type alias) for a generic class. I would like to do this so users of unit b can have access to TMyType without using unit a. I have units like this:

unit a;
interface
type
  TMyNormalObject = class
    FData: Integer;
  end;
  TMyType<T> = class
    FData: <T>;
  end;
implementation
end.

unit b;
interface
type
  TMyNormalObject = a.TMyNormalObject;  // works
  TMyType<T> = a.TMyType<T>; // E2508 type parameters not allowed on this type
implementation
end.

I already found a possible workaround, which I don't like because it can introduce hard to find bugs:

TMyType<T> = class(a.TMyType<T>); 

The problem with this approach is that it introduces a new class type and an a.TMyType instance is not a b.TMyType (while a.TMyNormallClass is a b.TMyNormalClass and vice versa - they are referring to the same class).

Sanson

Is it not possible for all users of TMyType<T> to refer from unit b? In that case your workaround might be feasible. – Thier

Why not create an interface and expose that to your users? – Pukka

Answer 1

It's currently not possible to declare a class type for a generic class.

See QC76605 for more information. Also the update below.

Example :

TMyClass<T> = class
end;
TMyClassClass<T> = class of TMyClass<T>; //E2508 type parameters not allowed on this type

The workaround that is presented looks like this :

TMyIntClass = TMyType<Integer>;
TMyIntClassClass = Class of TMyIntClass;

But as commented, that would defeat the whole idea of generics, since the class would have to be subclassed for every generic instantiation.

Here is also a link to a similar workaround on generating a specialized subclass of a generic type: derive-from-specialized-generic-types. In this case it would look like this :

TMySpecialClass = Class(TMyType<Integer>);

Update :

The workaround proposed by RM:

TMyType<T> = class(a.TMyType<T>);

can be implemented with type safety using following scheme:

unit Unita;
interface
type
  TMyType<T> = class
    Constructor Create;
  end;

implementation

uses
  Unitb;

constructor TMyType<T>.Create;
begin
  Inherited Create;
  //WriteLn( Self.QualifiedClassName,' ',Unitb.TMyType<T>.QualifiedClassName);
  Assert(Self.QualifiedClassName = Unitb.TMyType<T>.QualifiedClassName);
end;

end.

---

unit Unitb;

interface

uses Unita;

type
  TMyType<T> = class(Unita.TMyType<T>);
implementation
end.

---

Project Test;
{$APPTYPE CONSOLE}    
uses
  System.SysUtils,
  Unita in 'Unita.pas',
  Unitb in 'Unitb.pas';

var
  t1 : Unita.TMyType<Integer>;
  t2 : Unitb.TMyType<Integer>;
  t3 : TMyType<Integer>;    
begin
  try
    //t1 := Unita.TMyType<Integer>.Create;  //Exception EAssertionFailed !!
    t2 := Unitb.TMyType<Integer>.Create;
    t3 := TMyType<Integer>.Create;
    ReadLn;
  finally
    //t1.Free;
    t2.Free;
    t3.Free;
  end;
end.

When creating the generic class, a test is made to check that the created class is derived from the type declared in unit b. Thereby all attempts to create this class from unit a is detected.

Update 2:

Just to be clear, a reference to a generic class, "class of type<T>" is not possible, but a copy of a generic class is fine.

Thier

Hi LU RD, thanks but sorry, the question has been edited by Wharren and he changed the meaning of it. I changed it back. I would like to declare a type alias for the generic type and not a class type. I know that my workaround works with limitation. – Sanson

Ok, I understand. But to avoid any ambiguity you should rephrase "type alias" to "class reference". – Thier

Well I am looking for a type alias to a class and not a class reference. – Sanson

Well then, the compiler cannot resolve neither "type alias" nor "class references" for generic classes at the moment. Sorry for the name confusion. See if there is a possibility to use interfaces as proposed by whosrdaddy. – Thier

Answer 2

Since it is not possible to declare a "type alias" for a generic class, here is a solution using an interface.

unit UnitA;

interface  

Uses UnitB; 

type
  TMyType<T> = class(TInterfacedObject,ITMyType<T>)
    FData : T;
    Constructor Create( aV : T);
  end;

implementation

constructor TMyType<T>.Create( aV : T);
begin
  Inherited Create;
  FData := aV;
  WriteLn( Self.QualifiedClassName);
end;

end.

---

unit UnitB;

interface

type
  ITMyType<T> = Interface
  End;

implementation

end.

---

program Test;
{$APPTYPE CONSOLE}
uses
  UnitA in 'UnitA.pas',
  UnitB in 'UnitB.pas';

var
  it1 : ITMyType<Integer>;
begin
  it1:= TMyType<Integer>.Create(1);
  ReadLn;
end.

Thier

Answer 3

type  
    TMyArrOfT<T> = array of T;  
    TMyIntegers = TMyArrOfT<integer>;  

procedure x;  
var  
   t4: TMyIntegers;  
begin  
   t4 := [1,3,5,7,11];   //...
   for var i:integer := Low(t4) to High(t4) do  
      Memo1.Lines.Add(t4[i].ToString);  
end;  


type
    TMyArrOfT<T> = array of T;
    TMyButtons  = TMyArrOfT<TButton>;

procedure TForm1.Button1Click(Sender: TObject);
var
   t4: TMyButtons;
begin
   t4                 := [Button1, Button1, Button1, Button1, Button1]; // ...
   for var i: integer := Low(t4) to High(t4) do
      Memo1.Lines.Add(t4[i].Caption );
 end;

Goatskin

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